Take the mass of the Earth to be 5.98 ×
1024 kg.
If the Earth’s gravitational force causes a
falling 52 kg student to accelerate downward
at 9.8 m/s2, determine the upward accelera-
tion of the Earth during the student’s fall.
Answer in units of m/s2.
The same force acts on each mass
massstudent*g=massEarth*a
solve for a.
Oh, the Earth must be feeling a little left out, wanting to know its own acceleration, huh? Well, let's crunch some numbers and put a smile on the Earth's face!
We can use the formula for gravitational force (F = m*a) to find the force exerted on the student by the Earth. Since the student is falling, we can assume it's going downward, so the force will be positive. Using the given values, we have:
F = (52 kg) * (9.8 m/s^2) = 509.6 N
Now, to find the upward acceleration of the Earth, we can use Newton's third law: for every action, there is an equal and opposite reaction. The force exerted by the student on the Earth is the same as the force exerted by the Earth on the student. So, the acceleration of the Earth (a_Earth) can be found using the Earth's mass (m_Earth) and the force:
F = m_Earth * a_Earth
Rearranging the equation a bit, we have:
a_Earth = F / m_Earth
Plugging in the values, we get:
a_Earth = 509.6 N / (5.98 × 10^24 kg) ≈ 8.52 × 10^(-23) m/s^2
So, the Earth's upward acceleration during the student's fall is approximately 8.52 × 10^(-23) m/s^2. Just a tiny, itty-bitty acceleration for our grand ol' Earth. It's probably not even breaking a sweat!
To determine the upward acceleration of the Earth, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
1. Calculate the force exerted by the Earth on the student:
The force (F) exerted by the Earth on the student can be determined using Newton's second law of motion:
F = m * a
where:
m = mass of the student = 52 kg
a = acceleration of the student = 9.8 m/s^2
So, F = 52 kg * 9.8 m/s^2 = 509.6 N
2. The force exerted by the student on the Earth is equal in magnitude but opposite in direction to the force exerted by the Earth on the student:
Therefore, the force exerted by the student on the Earth is also 509.6 N.
3. Use Newton's second law to calculate the acceleration of the Earth (a_Earth):
F = m_Earth * a_Earth
where:
m_Earth = mass of the Earth = 5.98 × 10^24 kg
F = force exerted by the student on the Earth = 509.6 N
Rearranging the equation, we have:
a_Earth = F / m_Earth
a_Earth = 509.6 N / 5.98 × 10^24 kg
a_Earth ≈ 8.52 × 10^(-23) m/s^2
Therefore, the upward acceleration of the Earth during the student's fall is approximately 8.52 × 10^(-23) m/s^2.
To determine the upward acceleration of the Earth during the student's fall, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
In this case, the action is the gravitational force pulling the student downward, and the reaction is the force with which the student pulls the Earth upward. The magnitudes of these forces are equal, but their directions are opposite.
To find the upward acceleration of the Earth, we can use the equation:
Force = mass x acceleration
The mass of the student, given as 52 kg, remains the same. The acceleration of the student, given as 9.8 m/s^2 (meters per second squared), is the downward acceleration due to Earth's gravity.
Since the forces exerted by the student and the Earth are equal in magnitude, we can write:
Force of the student on the Earth = Force of Earth on the student
mass of student x acceleration of student = mass of Earth x acceleration of Earth
Plugging in the known values:
52 kg x 9.8 m/s^2 = 5.98 x 10^24 kg x acceleration of Earth
Now we can solve for the acceleration of the Earth:
acceleration of Earth = (52 kg x 9.8 m/s^2) / (5.98 x 10^24 kg)
Calculating this expression gives us the upward acceleration of the Earth during the student's fall.